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number of trees

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number of trees

by clock60 » Tue May 18, 2010 2:06 pm
in how many different ways can a gardener make a planting of 10 trees in 3 days if he must plant at least 1 tree in any given day

(A) 12
(B) 36
(C) 6
(D) 8
(E) 48

ao is coming soon
Last edited by clock60 on Tue May 18, 2010 10:41 pm, edited 1 time in total.
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by indiantiger » Tue May 18, 2010 4:03 pm
Choice A) and E) are the same ?
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by frank1 » Tue May 18, 2010 6:31 pm
May be C
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by liferocks » Tue May 18, 2010 8:06 pm
I am getting the ans as 10P3*3^7..this is not in option.

logic is he has to plant 3 trees in 3 day(one each) we can select and arrange 3 trees from 10 in 10P3 ways

for the remaining 7 trees each can be planted in any of the 3 days so to tal 3*3*3*3*3*3*3=3^7

So ans 10P3*3^7

Can any body please check what is wrong with this approach?
"If you don't know where you are going, any road will get you there."
Lewis Carroll
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by mj78ind » Tue May 18, 2010 10:01 pm
Do not know how to approach this question but Process Of Elimination is good friend!

I can readily make more than 12 combinations hence the answer is 36 ........
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by clock60 » Tue May 18, 2010 10:51 pm
hi guys
sorry for posting 2 identical answers, it was a typo and i corrected it
the answer to the problem -36
but i did not see any smart way to solve,
i actually count all number of possible outcomes ,after some day the combinations start to repeat
but for sure it is not the way to follow as prone to mistake and time consuming
one smart math student advise me to look at the problem as:
in how many way we can fill 3 boxes with 10 balls so that every box contain at least 1 box but any way not clear
i do not think this problem too hard the only we need is good approach
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by mj78ind » Tue May 18, 2010 11:12 pm
I do not know if this is the most efficient but it is an approach nonetheless.

Say he plants 1 tree in day one, now on day 2 and 3 he can plant 9 tress in 8 ways - 8,1;7,2;3,6;4,5;5,4;6,3;7,2;8,1

Now if he plants 2 trees in day one, now on day 2 and 3 he can plant 8 trees in 7 ways: 1,7;2,6;3,5;4,4;5,3;6,2;7,1

Thus, we can go all the way to 8 trees in one day and adding up ways of doing this is: 8+7+6+5+4+3+2+1 = 36 ways.

Hence, my answer is B
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by kstv » Tue May 18, 2010 11:44 pm
One tree has to be planted in each of the three days
Left with 10 -3 = 7 which can be planted in any of the three days.
is this is possible in 7 C 3 ways ??
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by clock60 » Wed May 19, 2010 2:13 am
hi
i was told how to solve this and want to share the way with others
the first step is the same as mentioned kstv
at first plant 1 tree in 1 day, and left with 10-3=7 trees
then let us write in the line 7 trees designate them with 0 and days as (___)
0 0 0 0 0 0 0 ___ ____
it means that 1 the first day gardener will plant 7 trees and nothing in remaining 2 days he plants nothing
or it may look like
0 0 ___ 0 0____ 0 0 0
1 st day 2 trees, 2-d day 2 trees and third day 3 trees
we can vary it
and the total number of outcomes will looks like
9!/(7!*2!)=36
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by kikor65 » Thu May 20, 2010 10:38 am
ID the permutation of three positive integers that when added = 10

I. 1 + 2 + 7
II. 1 + 3 + 6
III. 1 + 4 + 5
IV. 2 + 3 + 5
V. 2 + 4 + 4
VI. 3 + 3 + 4

Total 6 sets of 3

Permutation of sets of 3 = 3! = 3*2*1 = 6

Therefore, 6 * 6 = 36
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by jeffedwards » Thu May 20, 2010 11:34 am
clock60 wrote:hi
we can vary it
and the total number of outcomes will looks like
9!/(7!*2!)=36
will you explain to me how you got 9C2? just trying to understand
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by clock60 » Fri May 21, 2010 12:07 pm
jeffedwards wrote:
clock60 wrote:hi
we can vary it
and the total number of outcomes will looks like
9!/(7!*2!)=36
will you explain to me how you got 9C2? just trying to understand
(k+m-1)C(m)=(k+m-1)!/(m!(k-1)!)
i am not sure but it is combination with repetition
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by evsistr » Sat May 22, 2010 3:22 am
It's not mentioned, but all of 10 trees are thought to be identical (otherwise the number of ways obviously is greater than 48).

Let us suppose there is a box with two partitions, so we have three distinct parts. It may look like that:

\ **** | * | ***** /

where '\' and '/' are box' immovable walls, '*' is one of 10 trees, and '|' is a partition. Let us think that the gardener plants trees from the first part on the first day, trees from the second part on the second day, etc.

Now we have to obtain the number of ways to place 10 trees in 3 sub-boxes. But instead of moving trees from one part to another, we'll move the partitions in all ways possible. There're 9 possible places for 2 partitions, hence the overall number of ways to place them is a binomial coefficient 9C2 = (8*9) / 2! = 8*9 / 2 = 4*9 = 36.

Answer B.
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